# Thread: convergences subtruction question..

1. ## convergences subtruction question..

i am given {An} , {Bn} so
An->a
Bn-An->0

prove that Bn->a

??

2. Originally Posted by transgalactic
i am given {An} , {Bn} so
An->a
Bn-An->0

prove that Bn->a

??
Is this in $\mathbb{R}^n$ or in a generalized metric space? Ill assume its the former.

We know that by the definition of a convergent sequence that if $\xi_n\to\xi$ then $\forall{\varepsilon>0}$ there exists $N$ such that if $N\leqslant{n}$ then $d(\xi_n,\xi)<\varepsilon$

So we have that $\forall\varepsilon>0$ there exists a $N$ such that $N\leqslant{n}$ implies that $d\left(A_n,a\right)<\frac{\varepsilon}{2}$

And that $\forall\varepsilon>0$ there exists a $N_0$ such that if $N_0\leqslant{n}$ then $d\left(B_n-A_n,0\right)<\frac{\varepsilon}{2}$

Now it should be pretty apparent that since the metric on $\mathbb{R}^n$ is $\left|\bold{x}-\bold{y}\right|$ that $d\left(B_n-A_n,0\right)<\varepsilon\implies d\left(B_n,A_n\right)<\varepsilon$

So now consider

$d\left(B_n,a\right)

For $n>\max\left\{N,N_0\right\}$

3. i cant imagine where each group of numbers where e/2
where N0?

can you please make a line to describe on the axes
where each term located
??

4. Originally Posted by transgalactic
i cant imagine where each group of numbers where e/2
where N0?

can you please make a line to describe on the axes
where each term located
??
I'm sorry, I do not know what you mean? These are generic terms $\varepsilon$ and $\frac{\varepsilon}{2}$ are interchangable, they are both constants which you pick. You could have easily made both distances less than $\varepsilon$ resulting in [ $d\left(B_n,a\right)<2\varepsilon$ it gives the same result